Thank you for helping me with the proof. But above only prove |x|=|g^(-1)xg| if the order of x is finite. What if the order of x is infinite? The definition I was given is that if the order of x is infinite, then we write |x| = 0. I don't know how to show |g^(-1)xg| = 0.

Thank you for helping me with the proof. But above only prove |x|=|g^(-1)xg| if the order of x is finite. What if the order of x is infinite? The definition I was given is that if the order of x is infinite, then we write |x| = 0. I don't know how to show |g^(-1)xg| = 0.

If has finite order then so does by above.
Thus, they are both finite or infinite.
And if finite then they have the same order.